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2 years ago

Please help with math question

Mathematics

1 answer:

ELEN [110]2 years ago

7 0

Answer:

y = 4x² -3x +5

Step-by-step explanation:

It can be helpful to plot the points on a graph. The curve through them opens upward, so you know <em>the first two answer choices are incorrect</em>.

<em>Looking at the location of the vertex</em>

The points (-1, 12) and (2, 15) have nearly the same y-value, so the vertex of the parabola will be nearly halfway between those points, at about ...

... x = (-1 +2)/2 = 1/2

Because the point at (2, 15) has a slightly higher y-value, you know it is farther from the vertex, so the vertex will be at an x-value that is less than 1/2.

For quadratic ...

... y = ax² +bx +c

the vertex is found at

... x = -b/(2a)

For the 3rd choice, this is x = -(-4)/(2·3) = 2/3. This is higher than 1/2, so we can eliminate this choice, too.

Only the 4th choice remains. (It passes the vertex location test: x = -(-3)/(2·4) = 3/8 < 1/2.)

_____

There are some other ways to solve this problem, too.

try the given points in the given equations. The +5 as a constant of all of them tell you that all will work for x=0. Both of choices 3 and 4 work for x=-1, so we have to try x=2. Choice 3: 3·2² -4·2 +5 = 9, not 15. Choice 4: 4·2² -3·2 +5 = 15. So Choice 4 is the answer.
write equations for a, b and solve them. a(-1)² +b(-1) +5=12; a(2)² +b(2)+5=15. or a-b=7, 2a+b=5 in standard form. Adding these gives 3a=12, so a=4 and Choice 4 is the answer.
use a graphing calculator to write the quadratic regression formula from the three points. This result is shown in the attachment. The (a, b, c) = (4, -3, 5) correspond to Choice 4.

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2 years ago

Which transmition will change figure A into figure B in math

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Answer:

The figures informed are in different scales and mirrored in this way it will be necessary to rotate 180 about the origin, dilate by a scale factor of 2/3.

Step-by-step explanation:

What is figure rotation?

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1 year ago

Which is an algebraic expression for five more then Z?

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2 years ago

Richard has just been given an l0-question multiple-choice quiz in his history class. Each question has five answers, of which o

myrzilka [38]

Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is $p = \frac{1}{5} = 0.2$

10 questions.

This means that $n = 10$

A) What is the probability that he will answer all questions correctly?

This is $P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024$

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so $P(X = 0)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

Since $P(X = 0) = 0.1074$ , from item b.

$P(X \geq 1) = 1 - 0.1074 = 0.8926$

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264$

$P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055$

$P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008$

$P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001$

$P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0$

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328$

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

8 0

3 years ago